acceptance sampling attributes

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ACCEPTANCE SAMPLING 
 
 
 
 
Some approximations: 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Hyp(N, n, p) 
Bin(n, p) Po( λ ) 
n>10 
p<0.1 
1.0
N
n
< 
n>10 
p+
N
n
<0.1 
 
Ex. Suppose we have a large lot. To control the quality we pick 10 units randomly. If 
at most one of them is defect then the lot is accepted otherwise it is rejected. The 
fraction defective is p. Calculate the acceptance probability for 6 different values of p 
and use these to draw the corresponding OC Curve. 
 
Solution: Since the lot is big we approximate the number of defective units in the 
sample with a binomial distribution. 
 
 Pa = 91100 )p1(p1
10)p1(p
0
10
−⋅





+−⋅





 
 
We calculate the acceptance probability for p = 1%, 5%, 10%, 15%, 20% and 30%. 
p = 0.01 Pa(0.01) = 91100 99.001.01
1099.001.0
0
10
⋅





+⋅





 ≈ 0.996 
p = 0.05 Pa(0.05) = 91100 95.005.01
1095.005.0
0
10
⋅





+⋅





 ≈ 0.914 
p = 0.10 Pa(0.10) = 91100 90.010.01
1090.010.0
0
10
⋅





+⋅





 ≈ 0.736 
p = 0.15 Pa(0.15) = 91100 85.015.01
1085.015.0
0
10
⋅





+⋅





 ≈ 0.544 
p = 0.20 Pa(0.20) = 91100 80.020.01
1080.020.0
0
10
⋅





+⋅





 ≈ 0.376 
p = 0.30 Pa(0.30) = 91100 70.030.01
10
70.030.0
0
10
⋅





+⋅





 ≈ 0.149 
 
0.2 0.4 0.6 0.8 1
0.2
0.4
0.6
0.8
1
 
The slope will be different for different situations. 
Pa 
 p 
 
 
Double sampling plans 
 
 
 
Ex. Suppose that a lot contains 1000 units. We have decided to use the following 
double sampling plan: 
 
1) Pick 30 units randomly. 
 
- if all are correct then accept the lot, 
- if three or more are defective then reject the lot, 
- if one or two units are defective go to point 2. 
 
2) 60 new units are selected at random. 
 
- if the number of defectives totally in both samples add up to at most two then 
accept the lot, 
- if the number of defectives totally in both samples add up to three or more 
then reject the lot. 
 
Suppose that the lot contains 2% defective units. How big is the acceptance 
probability? 
 
Solution: In this example the acceptance probability is quite as easy to calculate 
exact. Since the conditions for both a binomial approximation (n/N = 30/1000 ≈ 0.03 
< 0.1) and a Poisson approximation are fulfilled we can use the approximation that 
feels easiest. 
 
In the solution that follows the binomial approximation is used: 
 
n1 = 30, n2 = 60, c1 = 0, c2 = 2, r1 = r2 = 3 
 
Let di denote the number of defectives in sample ”i”. 
 
This means that 
Sample 1 
if d1 = 0 accept the lot, 
if d1 ≥ 3 reject the lot, 
if d1 = 1 or 2 take a new sample. 
 
Sample 2 
if d1 + d2 ≤ 2 accept the lot 
if d1 + d2 ≥ 3 reject the lot 
 
 
 
 
The we will accept the lot in the following situations: 
 
 
Sample 1 Sample 2 
d1=0 
d1=1 d2=0 
d1=1 d2=1 
d1=2 d2=0 
 
This gives the acceptance probability: 
 
Pa = P(d1=0) + P(d1=1 ∩ d2=0) + P(d1=1 ∩ d2=1) + P(d1=2 ∩ d2=0) 
 
The probabilities that contain the intersection between the number of defectives in 
sample 1 and sample 2 are solved by using conditional probabilities. 
 
Accept the lot in sample 1: P(d1 = 0) = 300 98.002.00
30
⋅





 ≈ 0.5455 
 
Accept the lot in sample 2 when d1 = 1: 
 
In sample 1: Take 30 observations 
P(d1 = 1) = 291 98.002.01
30
⋅





 ≈ 0.28736 
In sample 2: Take 60 observations 
P(d2=0 | d1=1) = 600 98.002.00
60
⋅





 ≈ 0.2976 
P(d2=1 | d1=1) = 591 98.002.01
60
⋅





 ≈ 0.3644 
 
Accept the lot is sample 2 when d1 = 2: 
 
In sample 1: Take 30 observations 
P(d1 = 2) = 282 98.002.02
30
⋅





 ≈ 0.0988 
In sample 2: Take 60 observations 
P(d2=0 | d1=2) = 600 98.002.00
60
⋅





 ≈ 0.2976 
 
 
 
These calculations are put together: 
 
Pa = P(d1=0) + P(d2=0 | d1=1) ⋅ P(d1=1) + P(d2=1 | d1=1) ⋅ P(d1=1) + 
 + P(d2=0 | d1=2) ⋅ P(d1=2) = 
 
= 0.5455 + 0.2976 ⋅ 0.3340 + 0.3644 ⋅ 0.3340 + 0.2976 ⋅ 0.0988 ≈ 0.7960 
 
 
 
 
 
 
 
 
Double sampling plans with the OC curve going through the points 
(p1, 1 – α) and (p2, β) where α = 5% and β = 10% 
 
 
n1 = n2 
 
 
Sampling- 
plan no 
 
 
1
2
p
p
 
 
Acceptance 
number 
 
Approximate value of pn1 
when Pa = 
Approx. 
value of 
ASN(p)/n1 for 
p95 
 
1c 
 
2c 
 
0.95 
 
0.50 
 
0.10 
 
1 11.90 0 1 0.21 1.00 2.50 1.170 
2 7.54 1 2 0.52 1.82 3.92 1.081 
3 6.79 0 2 0.43 1.42 2.96 1.340 
4 5.39 1 3 0.76 2.11 4.11 1.169 
5 4.65 2 4 1.16 2.90 5.39 1.105 
6 4.25 1 4 1.04 2.50 4.42 1.274 
7 3.88 2 5 1.43 3.20 5.55 1.170 
 
2n1 = n2 
 
 
Sampling- 
plan no 
 
 
1
2
p
p
 
 
Acceptance 
number 
 
Approximate value of pn1 
when Pa = 
Approx. 
value of 
ASN(p)/n1 for 
p95 
 
1c 
 
2c 
 
0.95 
 
0.50 
 
0.10 
 
1 14.50 0 1 0.16 0.84 2.32 1.273 
2 8.07 0 2 0.30 1.07 2.42 1.511 
3 6.48 1 3 0.60 1.80 3.89 1.238 
4 5.39 0 3 0.49 1.35 2.64 1.771 
5 5.09 1 4 0.77 1.97 3.92 1.359 
6 4.31 0 4 0.68 1.64 2.93 1.985 
7 4.19 1 5 0.96 2.18 4.02 1.498 
 
 
 
 
 
 
Ex. In a factory you buy large lots of bolts. When the lots arrive to the factory the 
quality is controlled using the following double sampling plan. 
 
Pick 30 bolts at random. If all are correct then accept the lot. If 3 or more are 
defective then reject the lot. If the sample consists of one or two defective bolts then 
you pick another 50 units. If both samples sum up to two or less defectives then the 
lot is accepted. Otherwise it is rejected. 
 
Draw an OC curve for this sampling plan. 
 
 
Solution: The sampling plan can be summarized as 
 
n1 = 30, n2 = 50, c1 = 0, c2 = 2 and r1 = r2 = 3. 
 
 
Acceptance probabilities for the fraction defectives 0.01, 0.02, 0.05, 0.10 and 0.20 
are calculated: 
 
 
 
p 
 
Acceptance probabilities in sample 1 and 
sample 2 
 
 
Pa 
 
0.01 
Sample 1: 
 P(d1 = 0) = 300 99.001.00
30
⋅





 ≈ 0.7397 
Sample 2: 
 
 P(d1=1 … d2=0) + P(d1=1 … d2=1) + 
 + P(d1=2 … d2=0) = 
 
 = 
291500 99.001.0
1
3099.001.0
0
50
⋅





⋅⋅





+ 
 + 291491 99.001.0
1
3099.001.0
1
50
⋅





⋅⋅





 + 
 + 282500 99.001.0
2
3099.001.0
0
50
⋅





⋅⋅





≈ 0.2240 
 
 
 
0.7397 + 0.2240 = 
 
= 0.9637 
 
 
 p = 0.02 ⇒ Pa ≈ 0.827 p = 0.05 ⇒ Pa ≈ 0.329 
 
 p = 0.10 ⇒ Pa ≈ 0.048 p = 0.20 ⇒ Pa ≈ 0.001 
 
 
 
 
 
 
 
The OC curve will obtain the following appearance: 
 
0.025 0.05 0.075 0.1 0.125 0.15
0.2
0.4
0.6
0.8
1
 
Suppose you want to find a simple sampling plan with almost the same OC curve as 
the double samplingplan. Use the fraction defectives 0.05 and 0.20 to determine 
such a plan. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Pa 
p 
 
Ex. Determine the sequential sampling plan with an OC curve that goes through the 
points (p1, α) = (0.02, 0.05) och (p2, β) = (0.10, 0.10). 
 
 
 
Solution: 
The constants h1, h2 and s in the rejection line and the acceptance line are 
calculated. (The value of the constant K will be found in the denominator of all three 
constants. 
 
( )
( )




−
−
=
21
12
p1p
p1plnK = ))10.01(02.0
)02.01(10.0(ln
−⋅
−⋅
 = )
018.0
098.0(ln 
 
We use this expression in the equations for h1, h2 and s: 
 
h1 = K
)1(ln β
α−
 = 
)
018.0
098.0(ln
)
10.0
05.01(ln −
 ≈ 1.329 
 
h2 = K
)1(ln
α
β−
 = 
)
018.0
098.0(ln
)
05.0
10.01(ln −
 ≈ 1.7056 
 
s = 
K
)
p1
p1(ln
2
1
−
−
 = 
)
018.0
098.0(ln
)
10.01
02.01(ln
−
−
 ≈ 0.0503 
 
 
The acceptance line becomes d1 = -h1 + sn = -1.329 + 0.0503n 
 
The rejection line becomes d2 = h2 + sn = 1.7056 + 0.0503n 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
If we calculate ASN(p) for different values of the fraction defective , p, then we obtain 
the following ASN-curve. 
 
 
p 1 – B ASN(p) = 30 + 50 (1 – B) 
 
0.01 291 99.001.0
1
30
⋅





+ 282 99.001.0
2
30
⋅





 ≈ 0.2570 
 
30 + 50 ⋅ 0.2570 ≈ 42.85 
 
0.02 291 98.002.0
1
30
⋅





+ 282 98.002.0
2
30
⋅





 ≈ 0.4328 
 
30 + 50 ⋅ 0.4328 ≈ 51.64 
 
0.03 291 97.003.0
1
30
⋅





+ 282 97.003.0
2
30
⋅





 ≈ 0.5390 
 
30 + 50 ⋅ 0.5390 ≈ 56.95 
 
0.05 291 95.005.0
1
30
⋅





+ 282 95.005.0
2
30
⋅





 ≈ 0.5975 
 
30 + 50 ⋅ 0.5957 ≈ 59.88 
 
0.07 291 93.007.0
1
30
⋅





+ 282 93.007.0
2
30
⋅





 ≈ 0.5354 
 
30 + 50 ⋅ 0.5354 ≈ 56.77 
 
0.09 291 91.009.0
1
30
⋅





+ 282 91.009.0
2
30
⋅





 ≈ 0.4265 
 
30 + 50 ⋅ 0.4265 ≈ 51.32 
 
0.12 291 88.012.0
1
30
⋅





+ 282 88.012.0
2
30
⋅





 ≈ 0.3494 
 
30 + 50 ⋅ 0.3494 ≈ 47.47 
 
0.15 291 85.015.0
1
30
⋅





+ 282 85.015.0
2
30
⋅





 ≈ 0.1438 
 
30 + 50 ⋅ 0.1438 ≈ 37.19 
 
0.1 0.2 0.3 0.4 0.5
25
30
35
40
45
50
55
60
 
 
 
 
In an earlier example we saw that the OC-curve for this double sampling plan almost 
 
 
 
 
 
 
 
ASN for the simple 
sampling plan 
ASN for the double 
sampling plan 
p = 0.013 p = 0.114 
 
The values of mx and my are found in the following table. 
 
 
 
A Dodge & Roming table 
 
AOQL = )
N
1
n
1(ym − n
xp mm = 
 
where mp is the fraction value that gives the maximum value of AOQ(p) 
 
 
 
 
 
c mx my c mx my 
 
0 
 
1.00 
 
 0.3679 
 
11 
 
 8.82 
 
 7.233 
1 1.62 0.8400 12 9.59 7.948 
2 2.27 1.371 13 10.37 8.670 
3 2.95 1.942 14 11.15 9.398 
4 3.64 2.544 15 11.93 10.13 
5 4.35 3.168 16 12.72 10.88 
6 5.07 3.812 17 13.52 11.62 
7 5.80 4.472 18 14.31 12.37 
8 6.55 5.146 19 15.12 13.13 
9 7.30 5.831 20 15.92 13.89 
10 8.05 6.528